Question

Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers. (Refer to the figure below)

Answer 1

The angular momentum of the system will remain conserved as there is no external torque involved. We know,

Angular momentum = (Moment of Inertia) x (angular velocity)

Consider the merry go-round to be rotating with the child at a certain angular velocity. When the child jumps of, we have to view the angular velocity of the child with respect to the centre. If the velocity increases in the direction of angular velocity (d), the final momentum of the child increases and as the total momentum is conserved, the momentum of the merry go-round will reduce and hence the angular velocity reduces. This can also be understood in terms of force. To push forward, the child pushes the merry go-round backwards and hence provides a negative torque that retards it. Similarly, it would lead to increase in angular velocity of merry go-round when the child jumps backwards(b). Angular momentum of jumping child is negative and hence the angular momentum of merry go-round becomes more positive such that total is equal to initial angular momentum. In the other two cases (a and c) the motion is perpendicular to the tangential direction. Due to inertia, the child still has the same velocity wchich it had just before jumping. Hence angular momentum along the inital axis will not change. Hence, angular velocity remains the same.